Encuentre el Node cuya suma con X tiene el conjunto máximo de bits

Dado un árbol, y los pesos de todos los Nodes y un número entero x , la tarea es encontrar un Node i tal que peso[i] + x tenga el conjunto máximo de bits. Si dos o más Nodes tienen la misma cantidad de bits establecidos cuando se agregan con x , encuentre el que tiene el valor mínimo.
Ejemplos: 
 

Aporte: 
 

x = 15 
Salida: 4 
Node 1: setbits(5 + 15) = 2 
Node 2: setbits(10 + 15) = 3 
Node 3: setbits(11 + 15) = 3 
Node 4: setbits(8 + 15) = 4 
Node 5: conjunto de bits (6 + 15) = 3 
 

Enfoque: Realice dfs en el árbol y realice un seguimiento del Node cuya suma con x tiene el máximo de bits establecidos. Si dos o más Nodes tienen el mismo número de bits establecidos, elija el que tenga el número mínimo.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int maximum = INT_MIN, x, ans = INT_MAX;
 
vector<int> graph[100];
vector<int> weight(100);
 
// Function to perform dfs to find
// the maximum set bits value
void dfs(int node, int parent)
{
    // If current set bits value is greater than
    // the current maximum
    int a = __builtin_popcount(weight[node] + x);
    if (maximum < a) {
        maximum = a;
        ans = node;
    }
 
    // If count is equal to the maximum
    // then choose the node with minimum value
    else if (maximum == a)
        ans = min(ans, node);
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
    x = 15;
 
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
 
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
 
    dfs(1, 1);
 
    cout << ans;
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
static int maximum = Integer.MIN_VALUE, x, ans = Integer.MAX_VALUE;
 
static Vector<Vector<Integer>> graph = new Vector<Vector<Integer>>();
static Vector<Integer> weight = new Vector<Integer>();
 
//number of set bits
static int __builtin_popcount(int x)
{
    int c = 0;
    for(int i = 0; i < 60; i++)
    if(((x>>i)&1) != 0)c++;
     
    return c;
}
 
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
    // If current set bits value is greater than
    // the current maximum
    int a = __builtin_popcount(weight.get(node) + x);
    if (maximum < a)
    {
        maximum = a;
        ans = node;
    }
 
    // If count is equal to the maximum
    // then choose the node with minimum value
    else if (maximum == a)
        ans = Math.min(ans, node);
         
    for (int i = 0; i < graph.get(node).size(); i++)
    {
        if (graph.get(node).get(i) == parent)
            continue;
        dfs(graph.get(node).get(i), node);
    }
}
 
// Driver code
public static void main(String args[])
{
    x = 15;
 
    // Weights of the node
    weight.add(0);
    weight.add(5);
    weight.add(10);;
    weight.add(11);;
    weight.add(8);
    weight.add(6);
     
    for(int i = 0; i < 100; i++)
    graph.add(new Vector<Integer>());
 
    // Edges of the tree
    graph.get(1).add(2);
    graph.get(2).add(3);
    graph.get(2).add(4);
    graph.get(1).add(5);
 
    dfs(1, 1);
 
    System.out.println( ans);
}
}
 
// This code is contributed by Arnab Kundu

# Python implementation of the approach
from sys import maxsize
 
maximum, x, ans = -maxsize, None, maxsize
 
graph = [[] for i in range(100)]
weight = [0] * 100
 
# Function to perform dfs to find
# the maximum set bits value
def dfs(node, parent):
    global x, ans, graph, weight, maximum
 
    # If current set bits value is greater than
    # the current maximum
    a = bin(weight[node] + x).count('1')
 
    if maximum < a:
        maximum = a
        ans = node
 
    # If count is equal to the maximum
    # then choose the node with minimum value
    elif maximum == a:
        ans = min(ans, node)
 
    for to in graph[node]:
        if to == parent:
            continue
        dfs(to, node)
 
# Driver Code
if __name__ == "__main__":
 
    x = 15
 
    # Weights of the node
    weight[1] = 5
    weight[2] = 10
    weight[3] = 11
    weight[4] = 8
    weight[5] = 6
 
    # Edges of the tree
    graph[1].append(2)
    graph[2].append(3)
    graph[2].append(4)
    graph[1].append(5)
 
    dfs(1, 1)
 
    print(ans)
 
# This code is contributed by
# sanjeev2552

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
static int maximum = int.MinValue, x,ans = int.MaxValue;
 
static List<List<int>> graph = new List<List<int>>();
static List<int> weight = new List<int>();
 
// number of set bits
static int __builtin_popcount(int x)
{
    int c = 0;
    for(int i = 0; i < 60; i++)
    if(((x>>i)&1) != 0)c++;
     
    return c;
}
 
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
    // If current set bits value is greater than
    // the current maximum
    int a = __builtin_popcount(weight[node] + x);
    if (maximum < a)
    {
        maximum = a;
        ans = node;
    }
 
    // If count is equal to the maximum
    // then choose the node with minimum value
    else if (maximum == a)
        ans = Math.Min(ans, node);
         
    for (int i = 0; i < graph[node].Count; i++)
    {
        if (graph[node][i] == parent)
            continue;
        dfs(graph[node][i], node);
    }
}
 
// Driver code
public static void Main()
{
    x = 15;
 
    // Weights of the node
    weight.Add(0);
    weight.Add(5);
    weight.Add(10);
    weight.Add(11);;
    weight.Add(8);
    weight.Add(6);
     
    for(int i = 0; i < 100; i++)
    graph.Add(new List<int>());
 
    // Edges of the tree
    graph[1].Add(2);
    graph[2].Add(3);
    graph[2].Add(4);
    graph[1].Add(5);
 
    dfs(1, 1);
 
    Console.Write( ans);
}
}
 
// This code is contributed by mits

<script>
 
// Javascript implementation of the approach
     
    let maximum = Number.MIN_VALUE, x,
    ans = Number.MAX_VALUE;
     
    let graph = new Array(100);
    for(let i=0;i<100;i++)
    {
        graph[i]=[];
    }
     
    let weight = [];
     
    //number of set bits
    function __builtin_popcount(x)
    {
        let c = 0;
        for(let i = 0; i < 60; i++)
            if(((x>>i)&1) != 0)
                c++;
           
        return c;
    }
     
    // Function to perform dfs to find
   // the maximum value
    function dfs(node,parent)
    {
        // If current set bits value is greater than
        // the current maximum
        let a = __builtin_popcount(weight[node] + x);
        if (maximum < a)
        {
            maximum = a;
            ans = node;
        }
       
        // If count is equal to the maximum
        // then choose the node with minimum value
        else if (maximum == a)
            ans = Math.min(ans, node);
               
        for (let i = 0; i < graph[node].length; i++)
        {
            if (graph[node][i] == parent)
                continue;
            dfs(graph[node][i], node);
        }
    }
     
    // Driver code
     
    x = 15;
   
    // Weights of the node
    weight.push(0);
    weight.push(5);
    weight.push(10);;
    weight.push(11);;
    weight.push(8);
    weight.push(6);
       
   
    // Edges of the tree
    graph[1].push(2);
    graph[2].push(3);
    graph[2].push(4);
    graph[1].push(5);
   
    dfs(1, 1);
   
    document.write( ans);
     
    // This code is contributed by unknown2108
     
</script>

4

Análisis de Complejidad: 
 

• Complejidad temporal: O(N). 
  En dfs, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida a dfs es O(N) si hay un total de N Nodes en el árbol. Además, para procesar cada Node se usa la función builtin_popcount() que tiene una complejidad de O(c) donde c es una constante y dado que esta complejidad es constante, no afecta la complejidad temporal general. Por lo tanto, la complejidad del tiempo es O(N).
• Espacio Auxiliar : O(1). 
  No se requiere ningún espacio adicional, por lo que la complejidad del espacio es constante.